### Example Problem **Problem:** Pizza Shack usually seats 4.2 customers every 10 minutes during the lunch service. What is the likelihood it takes between 2 and 4 minutes for the next customer to arrive? **Explanation:** In this problem, we need to figure out the probability that the time until the next customer arrives falls between 2 and 4 minutes. Given the rate at which customers arrive (4.2 per 10 minutes), this kind of problem can be approached using the concepts of Poisson processes and exponential distributions. 1. **Average Rate of Customer Arrivals (λ):** - The average rate λ can be calculated as 4.2 customers per 10 minutes or 0.42 customers per minute. 2. **Exponential Distribution Function:** - For a continuous variable, the probability density function (PDF) for the exponential distribution is given by: \[ f(t) = λe^{-λt} \] where \( t \) is the time in minutes. 3. **Calculate the Probability:** - To determine the probability that the time \( t \) is between 2 and 4 minutes, we need to compute the cumulative distribution function (CDF) for these time points and find the difference: \[ P(2 \leq t \leq 4) = F(4) - F(2) \] where \( F(t) = 1 - e^{-λt} \). 4. **Applying the Values:** - Compute \( F(4) \) and \( F(2) \): \[ F(4) = 1 - e^{-0.42 \times 4} = 1 - e^{-1.68} \] \[ F(2) = 1 - e^{-0.42 \times 2} = 1 - e^{-0.84} \] - Now find the difference: \[ P(2 \leq t \leq 4) = (1 - e^{-1.68}) - (1 - e^{-0.84}) \] \[ P(2 \leq t \leq 4) = e^{-0.84} - e^{-1.68} \] By solving the above expression using a calculator, you would get the exact numerical likelihood value. **Note:** The example
### Example Problem **Problem:** Pizza Shack usually seats 4.2 customers every 10 minutes during the lunch service. What is the likelihood it takes between 2 and 4 minutes for the next customer to arrive? **Explanation:** In this problem, we need to figure out the probability that the time until the next customer arrives falls between 2 and 4 minutes. Given the rate at which customers arrive (4.2 per 10 minutes), this kind of problem can be approached using the concepts of Poisson processes and exponential distributions. 1. **Average Rate of Customer Arrivals (λ):** - The average rate λ can be calculated as 4.2 customers per 10 minutes or 0.42 customers per minute. 2. **Exponential Distribution Function:** - For a continuous variable, the probability density function (PDF) for the exponential distribution is given by: \[ f(t) = λe^{-λt} \] where \( t \) is the time in minutes. 3. **Calculate the Probability:** - To determine the probability that the time \( t \) is between 2 and 4 minutes, we need to compute the cumulative distribution function (CDF) for these time points and find the difference: \[ P(2 \leq t \leq 4) = F(4) - F(2) \] where \( F(t) = 1 - e^{-λt} \). 4. **Applying the Values:** - Compute \( F(4) \) and \( F(2) \): \[ F(4) = 1 - e^{-0.42 \times 4} = 1 - e^{-1.68} \] \[ F(2) = 1 - e^{-0.42 \times 2} = 1 - e^{-0.84} \] - Now find the difference: \[ P(2 \leq t \leq 4) = (1 - e^{-1.68}) - (1 - e^{-0.84}) \] \[ P(2 \leq t \leq 4) = e^{-0.84} - e^{-1.68} \] By solving the above expression using a calculator, you would get the exact numerical likelihood value. **Note:** The example
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter12: Probability
Section12.3: Conditional Probability; Independent Events; Bayes' Theorem
Problem 59E
Related questions
Question
Question 9
Please use simple probability rules
![### Example Problem
**Problem:**
Pizza Shack usually seats 4.2 customers every 10 minutes during the lunch service. What is the likelihood it takes between 2 and 4 minutes for the next customer to arrive?
**Explanation:**
In this problem, we need to figure out the probability that the time until the next customer arrives falls between 2 and 4 minutes. Given the rate at which customers arrive (4.2 per 10 minutes), this kind of problem can be approached using the concepts of Poisson processes and exponential distributions.
1. **Average Rate of Customer Arrivals (λ):**
- The average rate λ can be calculated as 4.2 customers per 10 minutes or 0.42 customers per minute.
2. **Exponential Distribution Function:**
- For a continuous variable, the probability density function (PDF) for the exponential distribution is given by:
\[ f(t) = λe^{-λt} \]
where \( t \) is the time in minutes.
3. **Calculate the Probability:**
- To determine the probability that the time \( t \) is between 2 and 4 minutes, we need to compute the cumulative distribution function (CDF) for these time points and find the difference:
\[ P(2 \leq t \leq 4) = F(4) - F(2) \]
where \( F(t) = 1 - e^{-λt} \).
4. **Applying the Values:**
- Compute \( F(4) \) and \( F(2) \):
\[ F(4) = 1 - e^{-0.42 \times 4} = 1 - e^{-1.68} \]
\[ F(2) = 1 - e^{-0.42 \times 2} = 1 - e^{-0.84} \]
- Now find the difference:
\[ P(2 \leq t \leq 4) = (1 - e^{-1.68}) - (1 - e^{-0.84}) \]
\[ P(2 \leq t \leq 4) = e^{-0.84} - e^{-1.68} \]
By solving the above expression using a calculator, you would get the exact numerical likelihood value.
**Note:** The example](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06a9a237-982c-465e-8a17-4be5fc663e3e%2Fbae60c06-a686-48f9-aaed-bd7a94d0cbe0%2F3fxjhso_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Example Problem
**Problem:**
Pizza Shack usually seats 4.2 customers every 10 minutes during the lunch service. What is the likelihood it takes between 2 and 4 minutes for the next customer to arrive?
**Explanation:**
In this problem, we need to figure out the probability that the time until the next customer arrives falls between 2 and 4 minutes. Given the rate at which customers arrive (4.2 per 10 minutes), this kind of problem can be approached using the concepts of Poisson processes and exponential distributions.
1. **Average Rate of Customer Arrivals (λ):**
- The average rate λ can be calculated as 4.2 customers per 10 minutes or 0.42 customers per minute.
2. **Exponential Distribution Function:**
- For a continuous variable, the probability density function (PDF) for the exponential distribution is given by:
\[ f(t) = λe^{-λt} \]
where \( t \) is the time in minutes.
3. **Calculate the Probability:**
- To determine the probability that the time \( t \) is between 2 and 4 minutes, we need to compute the cumulative distribution function (CDF) for these time points and find the difference:
\[ P(2 \leq t \leq 4) = F(4) - F(2) \]
where \( F(t) = 1 - e^{-λt} \).
4. **Applying the Values:**
- Compute \( F(4) \) and \( F(2) \):
\[ F(4) = 1 - e^{-0.42 \times 4} = 1 - e^{-1.68} \]
\[ F(2) = 1 - e^{-0.42 \times 2} = 1 - e^{-0.84} \]
- Now find the difference:
\[ P(2 \leq t \leq 4) = (1 - e^{-1.68}) - (1 - e^{-0.84}) \]
\[ P(2 \leq t \leq 4) = e^{-0.84} - e^{-1.68} \]
By solving the above expression using a calculator, you would get the exact numerical likelihood value.
**Note:** The example
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